Question: Is ${599957}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {599957}= &&{5}\cdot100000+ \\&&{9}\cdot10000+ \\&&{9}\cdot1000+ \\&&{9}\cdot100+ \\&&{5}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {599957}= &&{5}(99999+1)+ \\&&{9}(9999+1)+ \\&&{9}(999+1)+ \\&&{9}(99+1)+ \\&&{5}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {599957}= &&\gray{5\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {5}+{9}+{9}+{9}+{5}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${599957}$ is divisible by $3$ if ${ 5}+{9}+{9}+{9}+{5}+{7}$ is divisible by $3$ Add the digits of ${599957}$ $ {5}+{9}+{9}+{9}+{5}+{7} = {44} $ If ${44}$ is divisible by $3$ , then ${599957}$ must also be divisible by $3$ Add the digits of ${44}$ $ {4}+{4} = \color{#9D38BD}{8} $ If $\color{#9D38BD}{8}$ is divisible by $3$ , then ${44}$ must also be divisible by $3$ $\color{#9D38BD}{8}$ is not divisible by $3$, therefore ${599957}$ must not be divisible by $3$.